# Cauchy sequence proof.

by cragar
Tags: cauchy, proof, sequence
 P: 2,468 1. The problem statement, all variables and given/known data Assume $x_n$ and $y_n$ are Cauchy sequences. Give a direct argument that $x_n+y_n$ is Cauchy. That does not use the Cauchy criterion or the algebraic limit theorem. A sequence is Cauchy if for every $\epsilon>0$ there exists an $N\in \mathbb{N}$ such that whenever $m,n\geq N$ it follows that $|a_n-a_m|< \epsilon$ 3. The attempt at a solution Lets call $x_n+y_n=c_n$ now we want to show that $|c_m-c_n|< \epsilon$ Lets assume for the sake of contradiction that $c_m-c_n> \epsilon$ so we would have $|x_m+y_m-x_n-y_n|> \epsilon$ $x_m> \epsilon+y_n-y_m$ since $y_n>y_m$ and we know that $x_m< \epsilon$ so this is a contradiction and the original statement must be true.
Emeritus
PF Gold
P: 4,500
 Quote by cragar 1. The problem statement, all variables and given/known data Assume $x_n$ and $y_n$ are Cauchy sequences. Give a direct argument that $x_n+y_n$ is Cauchy. That does not use the Cauchy criterion or the algebraic limit theorem. A sequence is Cauchy if for every $\epsilon>0$ there exists an $N\in \mathbb{N}$ such that whenever $m,n\geq N$ it follows that $|a_n-a_m|< \epsilon$ 3. The attempt at a solution Lets call $x_n+y_n=c_n$ now we want to show that $|c_m-c_n|< \epsilon$
It's always good to be careful with wording. We want to show that if n and m are big enough, that this inequality holds.

 since $y_n>y_m$
This is probably not true, especially since you haven't even said what n and m are besides arbitrary numbers!

 and we know that $x_m< \epsilon$
This is also probably not true since there's no reason to think the limit is zero

To get the contradiction you're going to want to use the triangle inequality on |(xn-xm)+(yn-ym)|
 P: 2,468 ok thanks for your response. So I take $|(x_n-x_m)+(y_n-y_m)| \leq |x_n-x_m|+|y_n-y_m|$ lets assume that $|x_n-x_m|+|y_n-y_m| > \epsilon$ Im going to rewrite it as $A+B> \epsilon$ so now we have $A> \epsilon -B$ Can I just say this since we know that $A< \epsilon$ and $B< \epsilon$ since $\epsilon$ can be any number bigger than zero, then both of these values should be less than $\frac{\epsilon}{2}$ therefore $A+B< \epsilon$ I have a feeling my last step is not ok
Emeritus
 Quote by cragar Can I just say this since we know that $A< \epsilon$ and $B< \epsilon$ since $\epsilon$ can be any number bigger than zero, then both of these values should be less than $\frac{\epsilon}{2}$ therefore $A+B< \epsilon$